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Equations of motion for OCR A-level Physics

Equations of motion

This page covers the following topics:

1. Basic equation of motion
2. Basic equations for uniform acceleration
3. Equations for uniform acceleration
4. Deriving equations of motion
5. Projectile motion

For an object moving at a constant velocity, the distance it has travelled during a given time interval at a given speed can be calculated using the following formula: distance = speed ร— time. The instantaneous velocity of an object is its velocity at a specific point in time during its motion, whereas its average velocity is the average of all instantaneous velocities during the object's motion.

Basic equation of motion

The initial velocity, final velocity, acceleration and displacement of an object travelling at a uniform acceleration can be calculated using the following formula: vยฒ โˆ’ uยฒ = 2as.

Basic equations for uniform acceleration

An object's initial velocity, final velocity, time of motion, displacement and acceleration can all be related using the following formulas: s = (u + v)/2 ร— t, v = u + at and s = ut + atยฒ/2. The appropriate formula can be used in questions according to the information given in it. For example, if the time, initial and final velocities are known when displacement needs to be found, the first formula would be used.

Equations for uniform acceleration

The equations of motion can be derived using the most basic one, s = ut, and the definition of acceleration, a = (v โˆ’ u)/t. Rearranging the formula of acceleration gives v = u + at.

The average velocity of an object can be found by taking the the average of the initial and final velocity, (u + v)/2. Substituting this into the basic equation of motion gives s = (u + v)/2 ร— t.

The third equation of motion, s = ut + (1/2)atยฒ, can be found using the area under a velocity-time graph for an object with an initial velocity, u, and final velocity, v and substituting in the formula for acceleration.

The fourth equation of motion, vยฒ = uยฒ + 2as, can be found using s = ut. Substituting the average velocity, (u + v)/2, and rearranging for t from the first equation of motion, s = (u + v)/2 ร— (v โˆ’ u)/a. Rearranging gives 2as = (u + v)(v โˆ’ u). Expanding out gives the fourth equation of motion.

Deriving equations of motion

An object thrown into the air follows a parabolic path called a trajectory and is called a projectile. This object will move under the influence of gravity.

Since gravity acts downwards, it only affects the vertical velocity of the object. Equations of motion can be used to calculate the vertical component of velocity. The object travels upwards at a decreasing rate, until its velocity becomes 0 and it starts to accelerate towards the ground until it reaches it.

There are no horizontal forces acting on the projectile, since air resistance is usually neglected, so the horizontal component of velocity remains constant during the time of flight and the equation velocity = distance รท time can be used.

Projectile motion

1

A ball is dropped from a certain height above the ground. Given that gravitational acceleration is 9.8 m/sยฒ and that the ball lands on the ground at 7 m/s, calculate the height from which the ball is dropped.

Since the ball is dropped, u = 0.
vยฒ โˆ’ uยฒ = 2as
7ยฒ โˆ’ 0ยฒ = 2 ร— 9.8 ร— s
s = 49 รท 19.6 = 2.5 m

2.5 m

A ball is dropped from a certain height above the ground. Given that gravitational acceleration is 9.8 m/sยฒ and that the ball lands on the ground at 7 m/s, calculate the height from which the ball is dropped.

2

A car travels a distance of 20 m at a constant acceleration. Given that its initial velocity is 9 m/s and its final velocity is 12 m/s, calculate the time taken for the car to travel through this distance correct to 3 significant figures.

s = (u + v)/2 ร— t
20 = (9 + 12)/2 ร— t
t = 20/10.5 = 1.90 s

1.90 s

A car travels a distance of 20 m at a constant acceleration. Given that its initial velocity is 9 m/s and its final velocity is 12 m/s, calculate the time taken for the car to travel through this distance correct to 3 significant figures.

3

A ball is dropped from a table of height 1.3 m. It falls at a constant gravitational acceleration of 9.8 m/sยฒ. How long does it take for the ball to reach the ground? Give your answer correct to 3 significant figures.

Since the ball is dropped, u = 0.
s = ut + (1/2)atยฒ
1.3 = 0 + (1/2)(9.8)tยฒ
Rejecting the negative root since time cannot be negative.
t = โˆš(1.3/4.9) = 0.515 s

0.515 s

A ball is dropped from a table of height 1.3 m. It falls at a constant gravitational acceleration of 9.8 m/sยฒ. How long does it take for the ball to reach the ground? Give your answer correct to 3 significant figures.

4

How long must the motion of an object travelling at a constant speed of 14 m/s be to cover 65 m?

distance = speed ร— time
t = 65 รท 14 = 4.64 s (3 s. f.)

4.64 s

How long must the motion of an object travelling at a constant speed of 14 m/s be to cover 65 m?

5

Define instantaneous velocity.

The instantaneous velocity of an object is its velocity at a specific point in time.

velocity at a specific time

Define instantaneous velocity.

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